Evaluate [3]63³.

When taking the $n$-th root of a number raised to the $n$-th power, the root and the power cancel each other out.

$$\sqrt[3]{63^3} = 63$$ $$\boxed{63}$$

First, use the property $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$:

$$\sqrt{\frac{297}{64}} = \frac{\sqrt{297}}{\sqrt{64}}$$

Next, simplify the numerator $\sqrt{297}$. Find the largest perfect square factor of 297.
$297 = 9 \times 33$.
So, $\sqrt{297} = \sqrt{9 \times 33} = \sqrt{9} \times \sqrt{33} = 3\sqrt{33}$.
Now, simplify the denominator $\sqrt{64}$:

$$\sqrt{64} = 8$$

Combine the simplified numerator and denominator:

$$\frac{3\sqrt{33}}{8}$$ $$\boxed{\frac{3\sqrt{33}}{8}}$$

$\frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7}$.
When a base is multiplied by itself multiple times, it can be written in power form where the base is the number being multiplied and the exponent is the number of times it is multiplied.
The base is $\frac{5}{7}$ and it is multiplied 4 times.

$$\left(\frac{5}{7}\right)^4$$ $$\boxed{\left(\frac{5}{7}\right)^4}$$

When multiplying terms with the same base, add their exponents. The base is $\frac{3}{11}$.
Add the exponents: $-1 + 2 + 1 + (-4)$.

$$-1 + 2 + 1 - 4 = 1 + 1 - 4 = 2 - 4 = -2$$

So the expression simplifies to:

$$\left(\frac{3}{11}\right)^{-2}$$

To evaluate a term with a negative exponent, use the rule $a^{-n} = \frac{1}{a^n}$ or $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$:

$$\left(\frac{3}{11}\right)^{-2} = \left(\frac{11}{3}\right)^2$$

Now, apply the exponent to both the numerator and the denominator:

$$\frac{11^2}{3^2} = \frac{11 \times 11}{3 \times 3} = \frac{121}{9}$$ $$\boxed{\frac{121}{9}}$$